I need your advise on the following challenge I have an 8 volt 0.3amp solar panel which I would like to use to charge a small lead acid battery 12 volt no more than 10 AH.
Could you suggest a simple JT circuit that could be used to get this done.
Many thanks
Andrew
Well, I’ve been experimenting with many DC-DC converters, including my latest, a high current Joule Thief. I was successful at getting 1 volt at 2.5 amps from my power supply boosted up to 6.6 volts, but the circuit must use very a low resistance MOSFET to ‘do the heavy lifting’. And I’m still not sure if the circuit will be easy to replicate for the average experimenter.
I found a source for a DC-DC converter (step down, or higher to lower voltage) on eBay, for under 2 dollars apiece. There are other sellers such as this one that sell boost or step up types to convert lower to higher voltage. The shipping is free, so for a few dollars I can get a DC-DC converter that is 90 percent efficient and is already on a circuit board, so it is very difficult for me to justify building my own – the economics are just too compelling.
All I have to do is mount the circuit board in a small project box along with some jacks or terminal strips and a switch, and maybe a LED to tell when it’s on. I may also add some filter chokes and/or capacitors and/or ferrite EMI suppressor sleeves to prevent the switching frequencies from getting out and interfering with other stuff.
It looks like this one would serve Andrew’s purposes nicely with an efficiency of 90 percent. The cost is $5.50 U.S. or about 3 and a half Pounds in the U.K.
They also sell similar ones that are current limited, to drive LEDs. This may work for charging a SLA battery, however since a solar cell is inherently current limited, the converter itself may not need to have current limiting.
Update Apr 17 I received a reply from Andrew saying that these were too expensive. I find it hard to believe that he could make one for cheaper (see note at end), but he asked if this one (Youtube video) would do the job. I think it would but with some changes to protect it and prevent damage.
First off, notice hoe brightly the neon lamp lights up when he connects the circuit up. This neon lamp is connected from the collector to the emitter of the 2N3055. A 2N3055 is rated for a maximum collector voltage of 60 volts, and a neon lamp takes about 90 volts to light up. This is half again as much as the maximum for the 2N3055, and it is absolutely foolish to subject the transistor to such as high voltage. It will most likely be damaged.
People may think, how can he say that when we can see that it’s working. Well, they haven’t taken into consideration what is happening in the circuit. The typical Joule Thief coil is wound with the same number of turns on the primary and the feedback windings. Therefore if the primary has 90 volts on it, the feedback winding should also have 90 volts on it, but of the opposite polarity. So when the transistor shuts off, the high emitter to base voltage causes the emitter to base junction to break down, and this damages the transistor by permanently reducing its current gain.
The solution is simple: reduce the number of turns on the feedback winding so that the voltage is much less than the primary winding. The feedback winding should have no more than 5 or 6 volts negative voltage on the base.
Another simple solution that may be easier than rewinding the coil is to put some form of protection on the base to prevent the voltage from going more than 5 volts negative on the base. This can be a 5.1 volt zener diode or a LED. If the zener diode is used, then it must have a 1N4148 diode connected in series with it, to prevent the zener from shunting the base current away from the base. If this were a small signal transistor such as a 2N4401, the LED would be a good choice. But the 2N3055 is a power transistor, and the base current will be much higher than for a small transistor, so if the LED is used, it should be a high current LED, or else put several regular LEDs in parallel.
Also, I would eliminate the neon lamp, and replace it with a zener diode. However, the zener must be able to dissipate the power coming out of the Joule Thief. This could be a few watts in this case. So I would use three 1 watt, 6 volt zener diodes in series, across the JT output to the battery leads. If the battery leads were accidently reversed (stuff happens, and never underestimate the human tendency to do stupid things) then the zeners could be damaged, so it would be necessary to put a 1N4003 rectifier in series with the zeners.
The video showed the diode on the output as a 1N4007, which is a slow recovery diode made for power line frequencies. At Joule Thief frequencies, it has poor performance because of its slow recovery time. Instead, a UF4007 fast recovery rectifier should be used.
More about this circuit later. The guy who made the video, dodoshlodo, made it difficult to get a good look at the schematic, what with his slewing the camera around and putting his hand in the way, so I’ll try to see more of this excessively long 10 minute video when I have more time.
Note: As I said, I find it hard to believe that he can make a charger for less than what these converters cost ($5.50 U.S. or 3 and a half Pounds in U.K.). Further more, the typical Joule Thief efficiency is 40 to 70 percent (in this case most likely towards the bottom of that range). If the DC-DC converter I suggested is used, the efficiency would be 90 percent, and the charging time would be dramatically reduced, probably to half as long. This alone more than justifies the difference in cost, if any.
I thought about it, and then it occurred to me that perhaps Andrew is not just making these for a hobby; perhaps he is making them for his business, which would justify counting the pennies. Perhaps I was duped into helping someone design a circuit for his business. I give this help to others for use as a hobby, not for a business. Consequently, I think I may not be offering Andrew further help in the future.